Revision of Lecture Twenty-One

Transcription

1 Revson of Lecture Twenty-One FFT / IFFT most wdely found operatons n communcaton systems Important to know what are gong on nsde a FFT / IFFT algorthm Wth the ad of FFT / IFFT, ths lecture looks nto OFDM system and other multcarrer systems 275

2 OFDM Modem OFDM basc concepts recall: Let {S k } be the complex-valued symbol sequence (e.g. QAM symbols) transmtted at rate f s or symbol perod T s I have delberately used the captal letter S k to denote transmtted symbols (nstead of the usual small letter s k ) and there s a good reason for t Indeed transmtted symbols S k are now vewed as frequency -doman samples At the OFDM transmtter: durng the perod T = NT s, N symbols S, S 1, S N 1 are transmtted, and complex baseband OFDM sgnal durng perod T s therefore s(t) = N 1 k= S k e j2π k T t At the OFDM recever: the receved sgnal s multple by e j2π T nt and ntegrated over T to obtan S n, n =, 1, N 1 1 T Z T s(t)e j2π n T t d t = 1 T N 1 k= S k Z T e j2π k T t e j2π n T t d t = S n No one really mplements OFDM Modem ths way, as t needs N oscllators (N pars of hardware modulators/demodulators) 276

3 DFT/FFT Implementaton Sample complex-valued baseband sgnal s(t) N tmes durng a perod T:.e. t = m N T «N 1 m s m = s N T = S k e j2πkm N, m =, 1,, N 1 Ths s just IDFT formula multplyng by a factor N, thus one can vew k= Transmtted symbols {S n } N 1 n= as a set of N frequency samples (hence captal S) Baseband sgnal samples {s m } N 1 m= as a set of tme samples (hence small s) That s, from the set of N frequency samples to obtan the set of N tme samples va IDFT: s m = N IDFT {S k } N k=, m =, 1,, N 1 At recever, durng a perod T, the set of N transmtted symbols {S n } N 1 n= s recovered from the set of N tme samples {s m } N 1 m= usng DFT: S n = 1 {s N DFT m } N m=, n =, 1,, N 1 IDFT/DFT are of course mplemented by IFFT/FFT 277

4 OFDM Transcever OFDM transmtter/recever: The carrer modulaton and demodulaton s standard. Also clock recovery, not shown, s standard. New components are the cyclc extenson add and remove Let transmtted frequency S 1 S S S 1 SN 1 S/P S N 1 P/S R R R N 1 IFFT s s 1 s N 1 FFT r r 1 r N 1 P/S S/P add C.Ext. rem. C.Ext. frame be [S S 1 S N 1 ] and transmtted tme frame be [s s 1 s N 1 ] From dscrete Fourer theory {s m } N 1 m= {S n} n= N 1 1 cos(2 π f c t) cos(2 π f ct) Actually, from fnte N frequency samples, tme doman sgnal has nfnte duraton, but ths tme doman sgnal s perodc wth a perod of N samples 278

5 Cyclc Extenson Why cyclc extenson Thus f the channel s deal, at recever, from N tme samples {r k } N 1 k=, N frequency samples {R k } N 1 k=,.e. the transmtted symbols can be recovered va FFT If the channel s dspersve, say, the CIR length s N h T s, then the transmtted length of N tme symbols wll spread to a length of N h + N, and N frequency samples s nsuffcent A soluton s to add some dummy symbols to make t N + N h frequency samples or cyclc extenson An equvalent and more effcent alternatve s to add cyclc extenson n a transmtted tme frame Add cyclc extenson at transmtter: The last N h tme samples s coped back to the begnnng of the frame, and transmtted samples are N +N h s N N h N h s N 1 s cyclc copy N+N h N s N Nh sn 1 279

6 Cyclc Extenson (contnue) Remove cyclc extenson at recever: number the current frame as and the N + N h tmes samples as N h, N h + 1,, 1,, 1,, N 1 Let the CIR be h,, h Nh and gnore nose for smplcty, we have: r, Nh = h s,n Nh + h 1 s 1,N 1 + h 2 s 1,N h Nh s 1,N Nh r, Nh +1 = h s,n Nh +1 + h 1 s,n Nh + h 2 s 1,N h Nh s 1,N Nh +1. r, 1 = h s,n 1 + h 1 s,n 2 + h 2 s,n h Nh s 1,N 1 r, = h s, + h 1 s,n 1 + h 2 s,n h Nh s,n Nh. r,nh = h s,nh + h 1 s,nh 1 + h 2 s,nh h Nh s,. r,n 1 = h s,n 1 + h 1 s,n 2 + h 2 s,n h Nh s,n Nh 1 Inter-frame nterference: transmtted samples from the prevous ( 1)th frame spread nto the frst N h tme samples of frame. Thus, the frst N h tme samples are dscarded 28

7 Cyclc Extenson (contnue) Remanng N samples are used to generate R,k, k N 1 Notng the cyclc extenson: s 1 = s N 1,, s Nh = s N Nh, and the last N tme samples n the current frame can be wrtten as: r,n = N h j= The N frequency samples are obtaned va FFT: h j s,n (j mod N), n N 1 R,k = N 1 n= r,n e j2π n N k, k N 1 Usng we have e j2π N k = e j2πn 1+1 N k = e j2πn 2+2 N k =, R,k = N h n= h n e j2π N n N 1 k n= s,n e j2π n N k = H k S,k where {H k } are the DFTs of the CIR {h k }, call frequency doman channel transfer functons (FDCTFs) 281

8 Equalsaton n OFDM In general, R,k = H k S,k + n,k wth n,k beng a channel nose component The transmtted symbols are determned by passng R,k through a decson devce: ( ) R,k S,k = Detector, k N 1 H k Note that the ntersymbol nterference occurs n the receved tme samples r,k, but ths does not matter, as the DFT removes ths ISI Ths s a beauty of OFDM: equalsaton becomes very smple Equalsaton n OFDM nvolves to estmate the FDCTFs {H } N 1 = Wth the estmated channel {Ĥ} N 1 =, the estmated transmtted symbols are gven by ( ) R,k Ŝ,k = Detector, k N 1 Ĥ k 282

9 Mult-Carrer CDMA Map a dfferent chp of a spreadng sequence to an ndvdual OFDM subcarrer Each OFDM subcarrer has a data rate dentcal to orgnal nput data rate Multcarrer absorbs ncreased rate due to spreadng n a wder frequency band MC-CDMA transmtter: s b k : th user s kth bt s = [s s 1 s G 1 ]: th user s spreadng code Processng gan s G (subcarrer number s also G) b k s 1 s G 1 IFFT P/S LPF cos(2 f t) π c 283

10 Mult-Carrer CDMA (contnue) MC-CDMA recever: Let number of users be I, and {,1, I 1}; kth receved symbol (sample) for subcarrer l s r k,l = I 1 =1 H l b k s l + n k,l H l : frequency response of lth subcarrer (subchannel), n k,l : nose sample Decson varable d k estmatng b k s G 1 d k = l= s l g lr k,l for g l : recprocal of estmated H l d k Σ s g s 1 s G 1 g G 1 g 1 FFT S/P A/D LPF cos(2 π f t) c 284

11 MC-DS-CDMA Parallel transmsson of DS-CDMA sgnals usng OFDM structure R b : nput bt rate, N: number of subcarrers, G: processng gan MC-DS-CDMA transmtter for user : nformaton data rate s R b bps, after S/P rate s R b N bps, after spreadng rate s R b N G bps s s cos(2 π f t) c b k S/P s IFFT P/S LPF 285

12 Summary OFDM mplementaton wth FFT: transmtted complex symbols S k are frequency samples, and transmtted tme sgnal samples s m are the IDFT of S k Cyclc extenson: the channel wth CIR length N h wll spread the transmtted frame from length N to N + N h By employng cyclc extenson, the nter-frame nterference can be removed by smply dscard the frst N h receved tme samples Equalsaton n OFDM becomes automatc : DFT smply removes ISI n the receved sgnal samples, all requred are estmatng frequency doman channel transfer functons MC-CDMA and MC-DS-CDMA 286